Inductors in parallel combine like resistors in parallel: the result is smaller than the smallest one. Paralleling inductors also shares current and reduces resistance.
How to use the Parallel Inductors Calculator
- Calculate
- Target equivalent inductance
- Inductor 1
- Inductor 2
- Inductor 3
- Inductor 4
- Inductor 5
- Inductor 6
- Inductor 7
- Inductor 8
- Inductor 9
- Inductor 10
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Choose the mode and enter the inductances (leave unused fields at 0).
Inputs explained
| Field | What to enter |
|---|---|
| Calculate | Choose one: Equivalent inductance, Missing inductor |
| Target equivalent inductance | Units: mH, µH, nH, H |
| Inductor 1 | Units: mH, µH, nH, H |
| Inductor 2 | Units: mH, µH, nH, H |
| Inductor 3 | Units: mH, µH, nH, H |
| Inductor 4 | Units: mH, µH, nH, H |
| Inductor 5 | Units: mH, µH, nH, H |
| Inductor 6 | Units: mH, µH, nH, H |
| Inductor 7 | Units: mH, µH, nH, H |
| Inductor 8 | Units: mH, µH, nH, H |
| Inductor 9 | Units: mH, µH, nH, H |
| Inductor 10 | Units: mH, µH, nH, H |
What each result means
- Equivalent inductance = 1÷[S]
- Missing inductor = 1÷(1÷[Target equivalent inductance] – [S])
Example values
How the main result changes when Target equivalent inductance changes and everything else stays at the example values above:
| Target equivalent inductance (mH) | Equivalent inductance (mH) |
|---|---|
| 2.5 | 5.9977 |
| 3.75 | 5.9977 |
| 5 | 5.9977 |
| 6.25 | 5.9977 |
| 7.5 | 5.9977 |
| 10 | 5.9977 |
Formula
1/L_eq = 1/L₁ + 1/L₂ + … + 1/L_n · Missing: L_x = 1/(1/L_target − Σ1/L_i)
Worked example
Inputs — Calculate: Equivalent inductance; Inductor 1: 10 mH; Inductor 2: 22 mH; Inductor 3: 47 mH; Inductor 4: 0 mH; Inductor 5: 0 mH; Inductor 6: 0 mH; Inductor 7: 0 mH; Inductor 8: 0 mH; Inductor 9: 0 mH; Inductor 10: 0 mH.
Results — Equivalent inductance = 5.99768 mH.
Key terms
- Equivalent inductance
- Single inductance with the same effect.
Frequently asked questions
Is the parallel inductance smaller than each inductor?
Yes, without coupling.
What about two inductors?
L = L₁L₂/(L₁ + L₂).
Why might the missing value be negative?
The target is larger than the existing inductors allow.
Does coupling matter?
Yes – mutual inductance changes the result.